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(D) The work done in blowing a soap bubble is given by $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.

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$W(4)^{\frac{1}{3}}$

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(D) The work done in blowing a soap bubble is given by $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,$\Delta A = 2 	imes (4 \pi r^2) = 8 \pi r^2$.For a spherical bubble,the volume is $V = \frac{4}{3} \pi r^3$,which implies $r^3 = \frac{3V}{4\pi}$,or $r = (\frac{3V}{4\pi})^{\frac{1}{3}}$.Substituting $r$ into the area formula,we get $A = 8 \pi (\frac{3V}{4\pi})^{\frac{2}{3}}$,which shows that $A \propto V^{\frac{2}{3}}$.Therefore,the work done $W$ is proportional to $V^{\frac{2}{3}}$,i.e.,$W \propto V^{\frac{2}{3}}$.If the volume changes from $V$ to $2V$,the new work $W'$ is given by $\frac{W'}{W} = (\frac{2V}{V})^{\frac{2}{3}} = 2^{\frac{2}{3}} = (2^2)^{\frac{1}{3}} = 4^{\frac{1}{3}}$.Thus,$W' = W(4)^{\frac{1}{3}}$.

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a soap bubble of volume $2V$ will be

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